Let $R$ be a ring and $M$ be an $R$-module. The flat dimension of $M$ is the infimum over all lengths of flat resolutions of $M$. Usually, the flat dimension of $M$ is denoted by $\mathrm{fd}_R(M)$. For example, $\mathrm{fd}_{\mathbb{Z}}(\mathbb{Q}) = 0$. Since $\mathbb{Q}$ has projective dimension $1$, the flat dimension and projective dimension of a module can be different. Sometimes they can be the same: $\mathbb{Z}/n$ for $n$ a positive integer has the same flat and projective dimension as $\mathbb{Z}$-modules.

The weak dimension of a ring $R$ is defined to be $\mathrm{w.dim}(R) = \sup_{M} \{ \mathrm{fd}_R(M) \}$ where $M$ runs over all left $R$-modules. Due to the symmetric nature of the tensor product, we can also take the supremum over all right $R$-modules, in contrast to the asymmetric nature of global dimension.
With respect to the weak dimension, the simplest type of ring is one with $\mathrm{w.dim}(R) = 0$. In other words, every $R$-module is flat. If $R$ is a field or a semisimple ring then $\mathrm{w.dim}(R) = 0$. An arbitrary product of fields also has weak dimension zero, and such rings are not semisimple.

If $R$ is a local ring (a ring with a unique maximal left ideal) with $\mathrm{w.dim}(R)= 0$, then $R$ is actually a division ring! This follows from these observations that hold for any ring $R$ with $\mathrm{w.dim}(R) = 0$:

If every $R$-module is flat, then for every finitely generated ideal $I$, the quotient $R/I$ is projective. This is because $R/I$ is flat and finitely presented, and so $R/I$ is projective.

Because $R/I$ is projective for any f.g. ideal $I$, the exact sequence $0\to I\to R\to R/I\to 0$ splits and hence we see that any finitely generated ideal is actually a principal ideal generated by an idempotent.

For any $a\in R$, the right ideal $aR$ is generated by an idempotent $e$ so that there exists $r,s\in R$ such that $ar = e$ and $es = a$. Hence $ara = a$.

In other words, for every $a\in R$ there is an $r\in R$ such that $ara = a$. In general, if this condition holds, then we say that $R$ is von Neumann regular. Actually, $\mathrm{w.dim}(R) = 0$ is equivalent to $R$ being von Neumann regular, although we have only proved one direction here.

Now, suppose $R$ is a local ring with $\mathrm{w.dim}(R) = 0$. We wish to show that $R$ is a division ring. So fix an arbitrary $a\in R$ and choose an $r\in R$ such that $ara = a$. This implies that $a(ra – 1) = 0$. If $a$ is already invertible, there is nothing to prove. If $a$ is not invertible, then it lies in the maximal ideal of $R$ and hence $ra – 1$ is invertible, showing that $a = 0$.